The general solution to the second-order differential equation y′′+10y=0 is in the form y (x) = c1cosβx+c2sinβx. Find the value of β, where β>0.

Answer:β=√10 or 3.16 (rounded to 2 decimal places)

Step-by-step explanation:

To find the value of β:

we will differentiate the y (x) equation twice to get a second order differential equation. We compare our second order differential equation with the Second order differential equation specified in the problem to get the value of β

y (x) = c1cosβx+c2sinβx

we use the derivative of a sum rule to differentiate since we have an addition sign in our equation.

Also when differentiating Cosβx and Sinβx we should note that this involves function of a function. so we will differentiate βx in each case and multiply with the differential of c1cosx and c2sinx respectively.

lastly the differential of sinx = cosx and for cosx = - sinx.

Knowing all these we can proceed to solving the problem.

y=c1cosβx+c2sinβx

y' = β*c1*-sinβx+β*c2*cosβx

y'=-c1βsinβx+c2βcosβx

y''=β*-c1β*cosβx + (β*c2β*-sinβx)

y'' = - c1β²cosβx - c2β²sinβx

factorize - β²

y'' = - β² (c1cosβx + c2sinβx)

y (x) = c1cosβx+c2sinβx

therefore y'' = - β²y

y''+β²y=0

now we compare this with the second order D. E provided in the question

y''+10y=0

this means that β²y=10y

β²=10

B=√10 or 3.16 (2 d. p)